[python] Can Python test the membership of multiple values in a list?

I want to test if two or more values have membership on a list, but I'm getting an unexpected result:

>>> 'a','b' in ['b', 'a', 'foo', 'bar']
('a', True)

So, Can Python test the membership of multiple values at once in a list? What does that result mean?

I would say we can even leave those square brackets out.

array = ['b', 'a', 'foo', 'bar']
all([i in array for i in 'a', 'b'])

Both of the answers presented here will not handle repeated elements. For example, if you are testing whether [1,2,2] is a sublist of [1,2,3,4], both will return True. That may be what you mean to do, but I just wanted to clarify. If you want to return false for [1,2,2] in [1,2,3,4], you would need to sort both lists and check each item with a moving index on each list. Just a slightly more complicated for loop.

If you want to check all of your input matches,

>>> all(x in ['b', 'a', 'foo', 'bar'] for x in ['a', 'b'])

if you want to check at least one match,

>>> any(x in ['b', 'a', 'foo', 'bar'] for x in ['a', 'b'])

[x for x in ['a','b'] if x in ['b', 'a', 'foo', 'bar']]

The reason I think this is better than the chosen answer is that you really don't need to call the 'all()' function. Empty list evaluates to False in IF statements, non-empty list evaluates to True.

if [x for x in ['a','b'] if x in ['b', 'a', 'foo', 'bar']]:
    ...Do something...

Example:

>>> [x for x in ['a','b'] if x in ['b', 'a', 'foo', 'bar']]
['a', 'b']
>>> [x for x in ['G','F'] if x in ['b', 'a', 'foo', 'bar']]
[]

Another way to do it:

>>> set(['a','b']).issubset( ['b','a','foo','bar'] )
True

how can you be pythonic without lambdas! .. not to be taken seriously .. but this way works too:

orig_array = [ ..... ]
test_array = [ ... ]

filter(lambda x:x in test_array, orig_array) == test_array

leave out the end part if you want to test if any of the values are in the array:

filter(lambda x:x in test_array, orig_array)

Here's how I did it:

A = ['a','b','c']
B = ['c']
logic = [(x in B) for x in A]
if True in logic:
    do something

I'm pretty sure in is having higher precedence than , so your statement is being interpreted as 'a', ('b' in ['b' ...]), which then evaluates to 'a', True since 'b' is in the array.

See previous answer for how to do what you want.

The Python parser evaluated that statement as a tuple, where the first value was 'a', and the second value is the expression 'b' in ['b', 'a', 'foo', 'bar'] (which evaluates to True).

You can write a simple function do do what you want, though:

def all_in(candidates, sequence):
    for element in candidates:
        if element not in sequence:
            return False
    return True

And call it like:

>>> all_in(('a', 'b'), ['b', 'a', 'foo', 'bar'])
True